 Macrostate

A macrostate groups microstates sharing prescribed coarse data. The grouping depends on what is retained: here retain only total energy. For the two distinguishable two-level systems, the macrostates $E=0,\Delta ,2\Delta$ have state degeneracies $1,2,1$. Although two units are not macroscopic in size, they make this loss of information visible: knowing $E=\Delta$ does not tell us which unit is excited.

At inverse temperature $\beta \Delta =\ln 3$, each unit's excited-state probability is $p=1/4$. Their independence gives microstate probabilities $9/16,3/16,3/16,1/16$ in the order above. Adding within each macrostate givesThe total expected value of energy is therefore $\Delta (6/16)+2\Delta (1/16)=\Delta /2$, matching two units each with expected value of energy $\Delta /4$. Equal state degeneracies would give the wrong answer: macrostate probabilities combine counting with the Boltzmann factor. At very high positive temperature, all four microstates become equally likely, but these three macrostate probabilities approach $1/4,1/2,1/4$.